Explanation Given: The Equation are, x ′ ( t ) − 2 y ( t ) = 4 t , x ( 0 ) = 4 y ′ ( t ) + 2 y ( t ) − 4 x ( t ) = − 4 t − 2 , y ( 0 ) = − 5 Calculation: Consider the given Equation. x ′ ( t ) − 2 y ( t ) = 4 t , x ( 0 ) = 4... ( 1 ) y ′ ( t ) + 2 y ( t ) − 4 x ( t ) = − 4 t − 2 , y ( 0 ) = − 5... ( 2 ) The matrix form of Equation (1) and Equation (2) is, [ x ′ ( t ) y ′ ( t ) ] = [ 0 2 4 − 2 ] [ x ( t ) y ( t ) ] + [ 4 t − 4 t − 2 ] , x ( 0 ) = [ 4 − 5 ] ... ( 3 ) The characteristic Equation matrix [ 0 2 4 − 2 ] is, [ − r 2 4 − 2 − r ] − r [ 1 0 0 1 ] = 0 | − r 2 4 − 2 − r | = 0 r 2 ( 2 + r ) − 8 = 0 r = 2 , − 4 The Eigen vector for r = 2 is, X 1 ( t ) = e 2 t [ 1 1 ] The Eigen vector for r = − 4 is, X 2 ( t ) = e − 4 t [ − 1 2 ] The fundamental matrix for homogeneous system is, X ( t ) = [ X 1 ( t ) X 2 ( t ) ] ... ( 4 ) Substitute e − 4 t [ − 1 2 ] for X 2 ( t ) and e 2 t [ 1 1 ] for X 1 ( t ) in Equation (4). X ( t ) = [ e 2 t − e − 4 t e 2 t 2 e − 4 t ] ... ( 5 ) The inverse of Equation (5) is, X − 1 ( t ) = [ 2 3 e − 2 t 1 3 e − 2 t − 1 3 e 4 t 1 3 e 4 t ] The solution is given as, x ( t ) = X ( t ) X − 1 ( 0 ) x ( 0 ) + X ( t ) ∫ 0 t X − 1 ( s ) f ( s ) d s ... ( 6 ) Substitute [ x ( t ) y ( t ) ] for x ( t ) , [ 2 3 e − 2 t 1 3 e − 2 t − 1 3 e 4 t 1 3 e 4 t ] for X − 1 ( t ) , [ 4 s − 4 s − 2 ] for f ( s ) , [ e 2 t − e − 4 t e 2 t 2 e − 4 t ] for X ( t ) and [ 4 − 5 ] for x 0 in Equation (6)

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Chapter 9.7, Problem 24E

To determine

To find:

The solution of the initial value problem x′(t)−2y(t)=4t and y′(t)+2y(t)−4x(t)=−4t−2.

Expert Solution & Answer

Chapter 9 Solutions

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The solution of the initial value problem x ′ ( t ) − 2 y ( t ) = 4 t and y ′ ( t ) + 2 y ( t ) − 4 x ( t ) = − 4 t − 2 .

Solution Summary: The author explains that the matrix form of Equation (1) and (2) is left[cxprime

Videos

To find:

The solution of the initial value problem x′(t)−2y(t)=4t and y′(t)+2y(t)−4x(t)=−4t−2.

Chapter 9 Solutions

The solution of the initial value problem x ′ ( t ) − 2 y ( t ) = 4 t and y ′ ( t ) + 2 y ( t ) − 4 x ( t ) = − 4 t − 2 .

Solution Summary: The author explains that the matrix form of Equation (1) and (2) is left[cxprime

Videos

To find: The solution of the initial value problem x′(t)−2y(t)=4t and y′(t)+2y(t)−4x(t)=−4t−2.

Chapter 9 Solutions

To find:

The solution of the initial value problem x′(t)−2y(t)=4t and y′(t)+2y(t)−4x(t)=−4t−2.